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RD Sharma Solution for Class 9 Mathematics Chapter 21 - Surface Areas and Volume of a Sphere

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Our RD Sharma Textbook Solutions are considered extremely helpful for solving the tough questions which are asked in the CBSE Class 9 exam. TopperLearning Textbook Solutions are compiled by subject experts. Herein, you can find all the answers to the textbook questions for Chapter 21 - Surface Areas and Volume of a Sphere.

All our solutions are created in accordance with the latest CBSE syllabus, and they are amended from time to time to provide the most relevant answers. Our free RD Sharma Solutions for CBSE Class 9 Mathematics will strengthen your fundamentals in Mathematics and will help you in your attempts to score more marks in the final examination. CBSE Class 9 students can refer to our solutions any time — while doing their homework and while preparing for the exam.

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Exercise/Page

RD Sharma Solution for Class 9 Mathematics Chapter 21 - Surface Areas and Volume of a Sphere Page/Excercise 21.1

Solution 1

Solution 2

Solution 3

Solution 4

Solution 5

Solution 6

Solution 7

Solution 8



Solution 9

Solution 10


Solution 11



Solution 12



Solution 13

RD Sharma Solution for Class 9 Mathematics Chapter 21 - Surface Areas and Volume of a Sphere Page/Excercise 21.2

Solution 1

Solution 2

Solution 3

Solution 4

Solution 5

Solution 6

Solution 7

Solution 8


Solution 9


Solution 10

Solution 11

Solution 12

Solution 13

Solution 14

Solution 15

Solution 16

Solution 17

Solution 18

Solution 19

Solution 20

Solution 21

Solution 22

Inner radius (r1) of hemispherical tank  = 1 m
     Thickness of hemispherical tank       = 1 cm = 0.01 m
Outer radius (r2) of hemispherical tank = (1 + 0.01) m = 1.01 m Volume of iron used to make the tank  =                                                            

Solution 23

Radius (r) of capsule  
Volume of spherical capsule
                                       
Thus, approximately 22.46 mm3 of medicine is required to fill the capsule.

Solution 24

    Let diameter of earth be d. So, radius earth will be  .
    Then, diameter of moon will be  . So, radius of moon will be  .
    Volume of moon =    
    Volume of earth =   
      
    Thus, the volume of moon is  of volume of earth.

Solution 25

Solution 26

Solution 27

Solution 28

Solution 29

Solution 30

Solution 31

Solution 32

RD Sharma Solution for Class 9 Mathematics Chapter 21 - Surface Areas and Volume of a Sphere Page/Excercise 21.26

Solution 1

Sphere has only one surface i.e. curved surface, so number of faces = 1

Hence, correct option is (a).

Solution 2

A hemisphere has two surfaces: one top surface and other curved surface.

T.S.A. = 2∏r2 + (∏r2)   {Area of Top-face = ∏r2}

         = 3∏r2

Solution 3

begin mathsize 12px style Total space Surface space Area space of space Sphere space equals 4 πr squared space
Total space surface space Area space of space Hemisphere equals space 3 πr squared
therefore space Required space Ratio space equals space fraction numerator 4 πr squared space over denominator 3 πr squared space end fraction equals 4 over 3 equals space 4 space colon space 3
Hence comma space correct space option space is space left parenthesis straight d right parenthesis. end style

Solution 4

begin mathsize 12px style Height space of space sphere space equals space Diameter space equals space 2 straight r
Height space of space cube space equals space Side space of space cube space equals space Height space of space sphere space equals space 2 straight r
Volume space of space sphere space equals space 4 over 3 πr cubed
Volume space of space cube space equals space left parenthesis 2 straight r right parenthesis cubed space equals space 8 straight r cubed
Ratio space of space their space volumes space equals space fraction numerator begin display style 4 over 3 πr cubed end style over denominator 8 straight r cubed end fraction equals straight pi over 6 equals fraction numerator up diagonal strike 22 to the power of 11 over denominator 7 cross times up diagonal strike 6 subscript 3 end strike end fraction equals space 11 over 21 equals 11 space colon space 21
Hence comma space correct space option space is space left parenthesis straight d right parenthesis. end style

Solution 5

begin mathsize 12px style The space largest space sphere space that space can space be space cut space from space straight a space cube space of space side space 6 space cm space will space have space
its space diameter space equals space side space of space cube
straight i. straight e. space 2 straight r space equals space 6 space cm space rightwards double arrow space straight r space equals space 3 space cm
Volume space of space that space sphere space equals space 4 over 3 πr cubed space equals fraction numerator 4 over denominator up diagonal strike 3 end fraction straight pi cross times up diagonal strike 3 cross times 3 cross times 3 equals space 36 straight pi space cm cubed
Hence comma space correct space option space is space left parenthesis straight b right parenthesis. space end style

Solution 6

begin mathsize 12px style Volume space of space cylindrical space rod space equals space πr squared straight h
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space πr squared space left parenthesis 8 straight r right parenthesis space space space left square bracket straight h space equals space 8 straight r space left parenthesis given right parenthesis right square bracket
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 8 πr cubed
Now comma space if space spherical space balls space have space same space radius comma space then space the space volume space of space one space ball space equals space 4 over 3 πr cubed
therefore space No. space of space balls space equals space fraction numerator Volume space of space Cylindrical space Rod over denominator Volume space of space one space Rod end fraction equals fraction numerator 8 up diagonal strike πr cubed end strike over denominator begin display style 4 over 3 up diagonal strike πr cubed end strike end style end fraction equals space 6
Hence comma space correct space option space is space left parenthesis straight c right parenthesis. end style

Solution 7

begin mathsize 12px style Volume space of space sphere space equals space 4 over 3 πr cubed space equals space straight V
straight V subscript 1 over straight V subscript 2 equals fraction numerator up diagonal strike begin display style 4 over 3 straight pi end style end strike straight r subscript 1 superscript 3 over denominator up diagonal strike 4 over 3 straight pi end strike straight r subscript 2 superscript 3 end fraction equals fraction numerator straight r subscript 1 superscript 3 over denominator straight r subscript 2 superscript 3 end fraction equals 1 over 8
rightwards double arrow straight r subscript 1 over straight r subscript 2 equals 1 half
Now comma space Surface space Area space of space Sphere space equals space 4 πr squared space equals space straight S
straight S subscript 1 over straight S subscript 2 equals fraction numerator 4 πr subscript 1 superscript 2 over denominator 4 πr subscript 2 superscript 2 end fraction equals open parentheses straight r subscript 1 over straight r subscript 2 close parentheses squared space equals space open parentheses 1 half close parentheses squared space equals space 1 fourth space equals space 1 space colon space 4
Hence comma space correct space option space is space left parenthesis straight b right parenthesis. end style

Solution 8

begin mathsize 12px style Surface space Area space of space Sphere space rightwards double arrow space 4 πr squared space equals space 144 straight pi
rightwards double arrow straight r squared space equals space 36 space rightwards double arrow space straight r space equals space 6
Volume space of space Sphere equals 4 over 3 πr cubed space equals space 4 over 3 straight pi open parentheses 6 close parentheses cubed space equals space 288 straight pi space straight m cubed
Hence comma space correct space option space is space left parenthesis straight a right parenthesis. space end style

Solution 9

begin mathsize 12px style Volume space of space solid space sphere space equals 4 over 3 straight pi space open parentheses 10 close parentheses cubed space equals space fraction numerator 4000 straight pi over denominator 3 end fraction space cm cubed
Volume space 8 space solid space spheres space of space radius space left parenthesis say right parenthesis space straight r space equals space 8 space cross times 4 over 3 πr cubed equals fraction numerator 32 πr cubed over denominator 3 end fraction space cm cubed
Now comma space fraction numerator 32 up diagonal strike straight pi straight r cubed over denominator up diagonal strike 3 end fraction equals fraction numerator 4000 up diagonal strike straight pi over denominator up diagonal strike 3 end fraction
rightwards double arrow straight r space equals space open parentheses 1000 over 8 close parentheses to the power of bevelled 1 third end exponent space equals space 10 over 2 space equals space 5 space cm
Surface space Area space of space each space small space ball space equals space 4 πr squared space equals space 4 straight pi space open parentheses 5 close parentheses squared space equals 100 straight pi space cm squared
Hence comma space correct space option space is space left parenthesis straight a right parenthesis. end style

RD Sharma Solution for Class 9 Mathematics Chapter 21 - Surface Areas and Volume of a Sphere Page/Excercise 21.27

Solution 10

begin mathsize 12px style Edge space of space cube space equals space straight a
rightwards double arrow Volume space of space cube space equals space straight a cubed
If space Sphere space is space inscribed space inside space cube space then space straight a space equals space 2 straight r space rightwards double arrow straight r space equals space straight a over 2
Volume space of space sphere space equals space 4 over 3 πr cubed equals space 4 over 3 straight pi open parentheses straight a over 2 close parentheses cubed space equals space straight pi over 6 straight a cubed
Ratio space of space volume space of space sphere space to space volume space of space cube space equals space fraction numerator begin display style straight pi over 6 straight a cubed end style over denominator straight a cubed end fraction equals straight pi over 6
Hence comma space correct space option space is space left parenthesis straight d right parenthesis. end style

Solution 11

begin mathsize 12px style Volume space of space sphere space equals space 4 over 3 πr cubed
Sphere space casted space into space straight a space cone space of space height space straight r.
Let space the space radius space of space cone space equals space straight R
therefore space Volume space of space cone space equals space 1 third πR squared open parentheses straight r close parentheses
Volume space of space cone space equals space Volume space of space sphere
rightwards double arrow 1 third πR squared straight r space space equals space 4 over 3 πr cubed
rightwards double arrow straight R squared space equals space 4 straight r squared
rightwards double arrow straight R space equals space 2 straight r
Hence comma space correct space option space is space left parenthesis straight a right parenthesis. end style

Solution 12

Radius of sphere = r

Sphere touches cylinder at Top, Base and Lateral Surface.

Then,

2r = height of cylinder = h

r = Radius of cylinder

Volume of cylinder = ∏r2h

                         =∏r2(2r)

                         = 2∏r3

Hence, correct option is (c).


Solution 13

begin mathsize 12px style Volume space of space sphere space of space radius space straight r space equals space 4 over 3 πr cubed space equals space straight V subscript 1 space space space space space space space.... open parentheses 1 close parentheses
If space straight a space cylinder space is space circumscribing space the space sphere comma space then
diameter space of space cylinder space equals space diameter space of space sphere
height space of space cylinder space equals space diameter space of space sphere
rightwards double arrow Radius space of space cylinder space equals space Radius space of space sphere
Height space of space cylinder space equals space 2 straight r
Volume space of space cylinder space equals space straight V subscript 2 space equals space πr squared straight h
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space πr squared space left parenthesis 2 straight r right parenthesis
rightwards double arrow straight V subscript 2 equals space 2 πr cubed space space space space.... open parentheses 2 close parentheses
Dividing space equation space open parentheses 1 close parentheses space and space open parentheses 2 close parentheses
straight V subscript 1 over straight V subscript 2 equals fraction numerator begin display style bevelled 4 over 3 space πr cubed end style over denominator 2 πr cubed end fraction
rightwards double arrow straight V subscript 1 over straight V subscript 2 equals 2 over 3
Hence comma space correct space option space is space left parenthesis straight c right parenthesis. end style

Solution 14

begin mathsize 12px style Let space the space radius space of space cone space equals space straight r
Then comma space radius space of space hemisphere space equals space straight r space space space space left parenthesis Both space have space equal space bases right parenthesis
straight V subscript cone space equals space 1 third πr squared straight h
straight V subscript Hemisphere space equals 2 over 3 πr cubed
straight V subscript cone space equals space straight V subscript Hemisphere
rightwards double arrow fraction numerator 1 over denominator up diagonal strike 3 end fraction up diagonal strike straight pi straight r squared straight h space equals space fraction numerator 2 over denominator up diagonal strike 3 end fraction up diagonal strike straight pi straight r cubed space space space space space
rightwards double arrow straight h space equals space 2 straight r
rightwards double arrow straight h over straight r equals 2 over 1 space space space space open curly brackets straight r space equals space height space of space hemisphere close curly brackets
Hence comma space correct space option space is space left parenthesis straight b right parenthesis. end style

Solution 15

begin mathsize 12px style If space all space of space these space have space equal space bases comma space then space their space radii space are space equal.
Their space heights space are space same. space space space left parenthesis given right parenthesis
straight r space equals space straight h subscript 1 space equals space straight h subscript 2
straight V subscript cone space equals 1 third πr squared straight h subscript 1 equals 1 third πr squared open parentheses straight r close parentheses equals 1 third πr cubed
straight V subscript hemisphere equals space 2 over 3 πr cubed
straight V subscript cylinder space equals πr squared space straight h subscript 2 space equals space πr squared space open parentheses straight r close parentheses space equals πr cubed
straight V subscript cone space space colon space straight V subscript hemisphere space space colon space straight V subscript cylinder space equals 1 third space colon space 2 over 3 space colon space 1 space equals space 1 space colon space 2 space colon space 3
Hence comma space correct space option space is space left parenthesis straight a right parenthesis. end style

TopperLearning provides step-by-step solutions for each question in each chapter in the RD Sharma textbook for class 9. Access the CBSE Class 9 Mathematics Chapter 21 - Surface Areas and Volume of a Sphere for free. The textbook questions have been solved by our subject matter experts to help you understand how to answer them. Our RD Sharma Textbook Solutions will help you to study and revise, and you can easily clear your fundamentals of Chapter 21 - Surface Areas and Volume of a Sphere.

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